Coaching Centre
Surface area of pyramids and cones
Surface area of right pyramids using slant height, and the cone formula SA = πr² + πrs, with Pythagoras to find a missing slant height.
Pyramids
Add the base and the triangular faces. Each triangle's height is the slant height; find it with Pythagoras if you are given the vertical height.
A square pyramid has base $10$ cm and vertical height $12$ cm. Find its surface area.
Cones
$s$ is the slant height. $\pi rs$ is the curved surface (it unrolls into a sector).
A cone has radius $3$ cm and height $4$ cm. Find its surface area.
Square pyramid, base $10$, vertical height $12$. Surface area?
Slant height $13$.
Cone, radius $3$, height $4$. Surface area (2 d.p.)?
$24\pi$.
Slant height of a cone with radius $5$ and height $12$?
$\sqrt{169}$.
The exact curved surface area of a cone with $r = 5$, $s = 13$ is:
$\pi \times 5 \times 13$.
Exact total surface area of that cone? Type it like 90pi.
$25\pi + 65\pi$.
Square pyramid, base $6$, slant height $5$. Surface area?
$36 + 60$.
A cone has curved surface $60\pi$ and radius $6$. Find the slant height.
$6\pi s = 60\pi$.
Open cone (no base), radius $2$, height $5$. Curved surface area (2 d.p.)?
$s = \sqrt{29}$, $\pi \times 2 \times \sqrt{29}$.
New numbers every time. Keep going until you get five in a row.